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Tetrads in General Relativity

II. Tetrad and Vierbein fields

Vierbein and Metric

   The primitive object in the tetrad formalism is the vierbein field ${e_\mu }^a(x)$, in place of the spacetime metric in the coordinate formalism. The vierbeins are real 4 × 4 invertible matrices, with 16 independent components, acting as a soldering agent between the general manifold (Greek indices) and the Minkowski spacetime (Latin indices):

2.6

\[  {{\mathbf{g}}_\mu }(x) = {e_\mu }^a(x) {{\mathbf{e}}_a}(x){\quad}{e_\mu }^a(x): = {{\mathbf{g}}_\mu }(x) \cdot {{\mathbf{e}}^a}(x) \]

2.7

\[  {{\mathbf{e}}_a}(x) = {e^\mu }_a(x) {{\mathbf{g}}_\mu }(x){\quad}{e^\mu }_a(x): = {{\mathbf{g}}^\mu }(x) \cdot {{\mathbf{e}}_a}(x)  \]

   From their definition it follows that the local Lorentz frame vierbein fields ${e_\mu }^a(x)$ diagonalize the metric tensor as in (2.4). In terms of the inverse matrices ${e^\mu }_a(x)$, equation (2.1) becomes the so-called the inner product-signature constraint

2.8

\[{g_{\mu \nu }}(x){e^\mu }_a(x){e^\nu }_b(x) = {\eta _{ab}}{\qquad}\sqrt { - g} \det [{e^\mu }_a] = 1\]

This key result is basic to the use of orthonormal bases in curved spacetime,

   The above argument can be made more rigorous by recognizing that  the metric is a non-degenerate matrix. So, it can be diagonalized at each spacetime point through an orthogonal matrix: ${E^T}gE = {\text{diag}}({\lambda _0}, {\lambda _1},{\lambda _2},{\lambda _3})$, ${E^T}E = {\rm I}$. Re-scaling 

2.9

\[  {e^\mu }_a: = \frac{1}{{\sqrt {\left| {{\lambda _a}} \right|} }}{E^\mu }_a  \]

gives then  ${e^T}ge = {\text{diag}}(1, - 1, - 1, - 1) = \eta $. Since the metric signature $\eta$ is an intrinsic property of a manifold (Sylvester's law), this result establishes that a Lorentzian metric admits a Minkowskian base at each point of the manifold.